Problem: $\text E = \left[\begin{array}{rrr}1 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]$ and $\text A = \left[\begin{array}{rr}2 & 4 \\ 2 & -2 \\ -1 & 5\end{array}\right]$ Let $\text {H = EA}$. Find $\text H$. $ {H = }$
Explanation: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{A}$. $ \text {H}=\left[\begin{array}{rr}{1} & {-1} & {1} \\ 2 & 3 & 4\end{array}\right]\left[\begin{array}{rr} {2} & 4 \\ {2} & -2 \\ {-1} & 5\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(1,-1,1)\cdot(2,2,-1)\\\\ &=1 \cdot 2-1\cdot 2 + 1\cdot -1\\\\ &=-1 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1 \cdot 2 + 2\cdot 4 = 6$ (Choice B) B $1 \cdot 4-1\cdot -2 + 1\cdot 5 = 11$ (Choice C) C $2 \cdot 2+3\cdot 2 + 4\cdot -1 = 6$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}-1 & 11 \\ 6 & 22\end{array}\right]$